Is the Equation $ |tanx + cotx| = |tanx| + |cotx| $ True for Any Value of $x$?

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  • Thread starter DaalChawal
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In summary, the given statement $|tanx + cotx| = |tanx| + |cotx|$ is true for any value of $x$, except for values that are integer multiples of $\pi$ or $\frac{(2n+1)\pi}{2}$. This is because $tanx$ and $cotx$ always have the same sign, making the absolute values redundant. However, it is important to note that this is only valid for $x \neq 0$, as the statement does not hold true for $x=0$. Additionally, the Triangle Inequality states that $|a+b| \leq |a| + |b|$, which is a general rule and not equivalent
  • #1
DaalChawal
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Q. Is $ |tanx + cotx| = |tanx| + |cotx| $ true for any $x?$ If it is true, then find the values of $x$.

My Working -->

Since $tanx$ and $cotx$ always have the same sign, so this holds true for any value of $x$.
 
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  • #2
Okay I think this should hold true for any $x$ except $n \pi$ , $\frac{(2n+1) \pi }{2}$
Am I correct?
 
  • #3
In general, $\displaystyle \begin{align*} \left| a + b \right| \not\equiv \left| a \right| + \left| b \right| \end{align*}$, rather $\displaystyle \begin{align*} \left| a + b \right| \leq \left| a \right| + \left| b \right| \end{align*}$. That's called the Triangle Inequality.
 
  • #4
DaalChawal said:
Q. Is $ |tanx + cotx| = |tanx| + |cotx| $ true for any $x?$ If it is true, then find the values of $x$.

My Working -->

Since $tanx$ and $cotx$ always have the same sign, so this holds true for any value of $x$.
You have to be a bit more careful than this, as Prove It says but you essentially have \(\displaystyle \left | y + \dfrac{1}{y} \right | = |y| + \left | \dfrac{1}{y} \right |\), which is true for \(\displaystyle y \neq 0\).

So, yes.

-Dan
 

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